A circle, a square, and an equilateral triangle had the same area. If their perimeters represent their ages, who are the oldest and the youngest? 
Once upon a time, a circle, a square, and an equilateral triangle had the same area.
If their perimeters represent their ages, who are the oldest and the youngest?
Note: Perimeter of a circle is its circumference. Solve without a calculator.

Answer:
Let $s, r, x$ be the sidelength of the square, the radius of the circle, and the sidelength of the triangle, respectively.
$s^2 = \pi r^2 = x^2 \dfrac{ \sqrt{3}}{4}$
$s = r\sqrt{\pi} = x \sqrt{\dfrac{ \sqrt{3}}{4}} =  x {\dfrac{ \sqrt[4]{3}}{2}}$
Without using a calculator, still clearly, $s > r, s > x, x > r \Rightarrow s > x > r$.
But what I need is the comparison of $4s, 2\pi r,$ and $3x$.
Given $s > x,$ clearly $4s > 3s > 3x$.
editing this part below: it's still messy I'm on it right now.
Given $s > r$, clearly, $4s > 4r$,  $2 \pi r > 6r > 4r > 4s$.
Given $x > r,$ clearly, $$
okay, I can't seem to compare the others. I only know square's older than the triangle.
p.s. is it correct I tagged algebraic-geometry? lol
 A: Just do it.
For the square we have:
Area = $A= s^2$.  So $s = \sqrt A$.
Age = perimeter = $4s$.  So Age= $4\sqrt{A}$.
For the circle we have:
Area = $A = \pi r^2$ so $r = \frac 1{\sqrt \pi} \sqrt A$.
Age = circumference = $2\pi r$. So Age = $2\pi\cdot \frac 1{\sqrt \pi}\sqrt A = 2\sqrt \pi \sqrt A$.
For the triangle we have
Area = $A =x^2 \frac {\sqrt 3}{4}$ so $x =\frac 2{\sqrt[4]3}\sqrt A$
Age = perimeter = $3x$. So Age $= 3\cdot \frac 2{\sqrt[4]3}\sqrt A$.
We need to compare: $4\sqrt{A},2\sqrt \pi \sqrt A, 3\cdot \frac 2{\sqrt[4]3}\sqrt A$.  So...
Just do it.
Divide all by $\sqrt A$ to compare
$4, 2\sqrt \pi, 3\cdot \frac 2{\sqrt[4]3}$
Divide all by $2$ to compare
$2, \sqrt \pi, \frac {3}{\sqrt[4]3}$.
Square each to compare
$4, \pi, \frac {9}{\sqrt 3}= 3\sqrt 3$.
$\pi < 4$.  And to compare $4$ to $3\sqrt 3$ we can square and compare $16$ to $9\times 3=27$.  So $4 < 3\sqrt 3$.
And that's that.  As $\pi < 4$ and $4 < 3\sqrt 3$ we know by transitivity that $\pi < 3\sqrt 3$.
So circle has the least circumference, then square, then triangle.
....
Now here's the thing.... you should start developing an intuition for these things and you will learn that we can maximize area of a shape with a given parameter by make the shape more circular. Eventually it should become intuitive that given the same parameters a regular triangle will have less area than a square which will have less area than a regular pentagon which will have less area than a hexagon and a circle will have the most area per perimeter.
And therefore if a circle, square and triangle have the same area it should be intuitive that the circle can achieve that area with the least circumference and the triangle will require the most perimeter.
However no-one is born with that intuition. But it is one all mathematicians eventually develop.  (And it comes from doing problems like these.)
A: Given: $$s=r\sqrt{\pi}=x\cdot\frac{\sqrt[4]{3}}{2}$$
Note that $4s=4r\sqrt{\pi}\gt2\pi r$ if and only if $2\gt\sqrt{\pi}$, by cross-dividing. However as $\pi\lt4$, this inequality is clear. Thus the square has larger perimeter than the circle.
$4s=4x\cdot\frac{\sqrt[4]{3}}{2}\lt3x$ if and only if $\sqrt[4]{3}\lt\frac{3}{2}$, again by cross-dividing. This inequality is also true: can you see why?
So we conclude the square has a smaller perimeter than the triangle. The triangle is oldest and the circle is youngest.
PS I believe it is a general fact about (regular?) convex polyhedra that as the number of sides increases, the constant of proportionality (considered by continuous enlargements of the shape) of area $\propto$ perimeter will decrease.
A: For a given area a circle has a minimum perimeter among all regular polygons found by methods of calculus of variations ( converse of iso-perimetric problem, can be termed iso-areal for changed objective/constraint) .
A Circle (radius $r$) has an infinite number of sides, $a\to 0$
Accordingly any regular polygon of same area of (side length $a$ )  has more perimeter length than the Circle. Perimeters (diminishing ages) are
For a "diangle" area $ \pi r^2, a \to (\infty \; r) $
For a triangle area $ \pi r^2 = \dfrac{\sqrt3}{4} a^2\to3 a\approx =8.08\; r$
For a square area $ \pi r^2  =4 a^2 \to 4a \approx 7.2 r\;$
For a Circle area $ \pi r^2  \text { minimum perimeter } a=2 \pi \approx 6.28 \;r.$
